Free Nonplussed! by Julian Havil

nets.
    Table 9.2. Possibilities when the Rowett dice are thrown twice.

    Table 9.3. The Rowett dice compared again.

    To add to the confusion, alter the game to one in which one of the dice is chosen by each player and thrown twice with the winner the person with the higher total. Table 9.2 gives the three possible totals for each die and the frequency with which each occurs.
    Now, if we perform the calculations as before, we arrive at table 9.3 , which shows the arrows of dominance are reversed. That is, A → C → B → A and this time with probabilitiesrespectively.
    Table 9.4. Effron’s dice compared.

    Figure 9.7. Effron’s dice nets 1.

    Figure 9.8. Effron’s dice nets 2.

    Figure 9.9. Effron’s dice nets 3.
    Effron’s Dice
    Bradley Effron, a statistician at Stanford University, extended the idea to four dice, giving the specification for three such sets, as shown in figures 9.7 – 9.9 . In each case, A → B → C → D → A.
    This is a little more subtle since the possibility of draws (rethrows) exists and we will take the trouble to compile the tables, shown as table 9.4 . In each table, the event of matching numbers is represented by an X.
    The ambiguity is exactly the same in all four cases and we can representatively deal with just the first, with A competing withB. If we write p for the probability that A wins, we have

    Table 9.5. Coin-tossing comparisons.

    which makesand so A → B with a probability ofand, of course, the probability is the same for the other pairings.
    Coin Tossing
    The second type of nontransitive effect that we will consider involves the spin of a fair coin. Inevitably, Martin Gardner has considered it, but the author first came across the phenomenon in the Warwick University mathematics magazine Manifold , which has long since disappeared. Player A takes a fair coin and repeatedly spins it but before doing so asks player B to choose a sequence of three heads and tails, for example, HTH. Having done so, A chooses his own sequence. The coin is repeatedly spun until one of the two sequences appears: whosoever’s sequence it is, wins. There are only eight possible choices for the triplet and B might reasonably think that somewhere among them there is a best choice, but there isn’t.

    Figure 9.10. The initial tree diagram.
    The left-hand column of table 9.5 shows the eight possible choices that B can make. The middle column shows the corresponding choice that A should make in each case. If he makes that choice, the right-hand column shows the probability of A winning.
    Compiling the column of odds takes a little effort and makes ample use of tree diagrams. We will consider the essentially different pairings separately, dealing in detail with the first of them.

    The first three tosses could be HHH, in which case B wins. Otherwise, a tail will appear among them and if this is the case, no matter how many more tails appear, A needs two heads and B still needs all three heads to win. A will assuredly get his two heads before B gets the three and it is just a mater of time before A wins. So, of the eight possibilities for the first three tosses, A will win on seven of them and so the probability of A winning is
    This is easy. Now to a slightly harder case.

    Figure 9.11. The tree diagram pruned.

    Figure 9.12. The initial tree diagram.

    We will analyse the situation using a three-stage tree diagram, as shown in figure 9.10 .
    Figure 9.10 shows the two paths along which either A or B clearly wins, but a little thought allows us to prune the tree diagram. First, if the first toss is a T, no matter what happens subsequently, HH is needed to complete the sequence and these willonly start the sequence chosen by B; A must win; this means that the whole right-hand side of the tree diagram is a win for A. Further, if the first two tosses are HT, then A must win for the same reason. Lastly, if the first two tosses are HH, then B must win. The tree diagram reduces to figure 9.11 .

    Figure 9.13.